Michele_Laino
  • Michele_Laino
Tutorial on the Quaternions of Hamilton
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Michele_Laino
  • Michele_Laino
Astrophysics
  • Astrophysics
Thank you Michele, I will give it a read! :)
Michele_Laino
  • Michele_Laino
thanks!! :) @Astrophysics

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oleg3321
  • oleg3321
is this some sort of new language??? lol
Michele_Laino
  • Michele_Laino
no, the quaternions were introduced by W. R. Hamilton in the far 1833 :) lol
oleg3321
  • oleg3321
lol looks very complicated
Michele_Laino
  • Michele_Laino
no, they are not complicated, please study my definitions first @oleg3321
Astrophysics
  • Astrophysics
Very easy to understand, thanks again!
Michele_Laino
  • Michele_Laino
:) @Astrophysics
welshfella
  • welshfella
Interesting article. Well explained. Hamilton was Ireland's most famous mathematician . Unfortunately his life was cut short by alcoholism.
Michele_Laino
  • Michele_Laino
Thanks! @welshfella
DanJS
  • DanJS
I remember my calc 3 teacher mentioning those to look at if you got bored.. thanks @Michele_Laino
Michele_Laino
  • Michele_Laino
thanks! :) @DanJS
arindameducationusc
  • arindameducationusc
Thank you @Michele_Laino . I will definitely give it a read
Michele_Laino
  • Michele_Laino
Thanks! :) @arindameducationusc
anonymous
  • anonymous
quaternions aren't a field, they are a division ring
Michele_Laino
  • Michele_Laino
Quaternions are a field @oldrin.bataku
anonymous
  • anonymous
also you should introduce those other basis matrices \(\mathbf i,\mathbf k\) more naturally -- something like this: consider for \(z=a+bi\) we have: $$\begin{align*}a+bi\mapsto \begin{pmatrix}a+bi&0\\0&a-bi\end{pmatrix}&=\begin{pmatrix}a&0\\0&a\end{pmatrix}+\begin{pmatrix}bi&0\\0&-bi\end{pmatrix}\\&=a\begin{pmatrix}1&0\\0&1\end{pmatrix}+b\begin{pmatrix}i&0\\0&-i\end{pmatrix}\\&=a\cdot\mathbf1+b\cdot\mathbf i\end{align*}$$ and then we have that \(\mathbf i\cdot\mathbf j=\mathbf k\)
anonymous
  • anonymous
https://en.wikipedia.org/wiki/Quaternion#Noncommutativity_of_multiplication >The fact that quaternion multiplication is not commutative makes the quaternions an often-cited example of a strictly skew field.
Michele_Laino
  • Michele_Laino
I wrote that formula: \[{\mathbf{ij}} = - {\mathbf{ji}} = {\mathbf{k}}\]
anonymous
  • anonymous
https://en.wikipedia.org/wiki/Division_ring >In abstract algebra, a division ring, also called a skew field, is a ring in which division is possible. >Division rings differ from fields only in that their multiplication is not required to be commutative. ... Historically, division rings were sometimes referred to as fields, while fields were called “commutative fields”.
anonymous
  • anonymous
yes, I know you did, but you introduced \(\mathbf k\) as a matrix first without any motivation. the reason for its existence is that we have four linearly independent matrices \(\mathbf {1,i,j,ij}\) since the multiplication is bilinear, and then we just call \(\mathbf {ij}=\mathbf k\)
Michele_Laino
  • Michele_Laino
I never worked with non-commutative algebra
Michele_Laino
  • Michele_Laino
I meant I never cited non-commutative algebra
anonymous
  • anonymous
? fields have commutative multiplication by definition, whereas the quaternions \(\mathbb H\) are a non-commutative algebra over \(\mathbb R\) and a division ring, not a field
Michele_Laino
  • Michele_Laino
when I wrote a matrix in M2(C) it is supposed that the multiplication is non-commutative since as you know, the multiplication betwen matrices is non-commutative
anonymous
  • anonymous
lol, yes, clearly, and these matrices do not commute, so they cannot form a field. the quaternions are not a field.
zzr0ck3r
  • zzr0ck3r
:) right
zzr0ck3r
  • zzr0ck3r
For real man, tuts that are wrong are even worse. This is all simple definition stuff you are getting wrong...Please remove.
Michele_Laino
  • Michele_Laino
More precisely, I have used the term "field", since it is commonly used in mathematics literature, a more appropriate term can be "sfield" which indicate a non-commutative field. I have supposed that, being the non-commutative characteristic evident, there was no need to use the term "sfield"
anonymous
  • anonymous
field is not commonly used in mathematics literature for a skew field, this isn't the 1800s anymore
anonymous
  • anonymous
it's called a division ring, that is the appropriate, correct, and commonly used term for algebraic structures like this

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